Problem: If $x \veebar y = x-3y$ and $x \star y = 3x^{2}+y^{2}$, find $0 \veebar (2 \star -1)$.
Answer: First, find $2 \star -1$ $ 2 \star -1 = 3(2^{2})+(-1)^{2}$ $ \hphantom{2 \star -1} = 13$ Now, find $0 \veebar 13$ $ 0 \veebar 13 = 0-(3)(13)$ $ \hphantom{0 \veebar 13} = -39$.